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\title{\heiti\zihao{2} 习题11.2}
\author{中书君}
\date{\songti \today}

\begin{document}
\maketitle
\section{设 $\sum\limits_{n=1}^{\infty} y_{n}$ 是在正项级数 $\sum\limits_{n=1}^{\infty} x_{n}$ 中添加括号得到的一个新级数. 证明: 当级数 $\sum\limits_{n=1}^{\infty} y_{n}$ 收敛时,必有 $\sum\limits_{n=1}^{\infty} x_{n}$ 也收敛.}
\textbf{证}\quad
记$S_{n}=\sum\limits_{k=1}^{n} x_{k}$,$T_{n}=\sum\limits_{k=1}^{n} y_{k}$,则显然$T_{n}$是$S_{n}$的一个子列,记$T_{k}=S_{n_{k}}$,则由于级数为正项技级数,故部分和序列单调.从而$T_{k}\geqslant S_{m}(m<n_{k}).$

又因为$T_{n}$有上界,从而$S_{n}$有上界,由单调有界定理知$\lim\limits_{n\rightarrow \infty}S_{n}=\sum\limits_{n=1}^{\infty} x_{n}$收敛.

\section{讨论下列级数的敛散性}
\subsection{ $\sum\limits_{n=1}^{\infty} \frac{3 n}{n^{3}+1}$}
\textbf{解}\quad
$$
	\begin{aligned}
		\sum\limits_{n=1}^{\infty} \frac{3 n}{n^{3}+1} & <\sum\limits_{n=1}^{\infty} \frac{3 n}{n^{3}}=\sum\limits_{n=1}^{\infty} \frac{3 n}{n^{3}+1} \\
		                                        & =\sum\limits_{n=1}^{\infty} \frac{3}{n^{2}}
	\end{aligned}
$$
收敛,由比较判别法知$\sum\limits_{n=1}^{\infty} \frac{3 n}{n^{3}+1}$收敛.

\subsection{ $\sum\limits_{n=1}^{\infty} \frac{1}{n 2^{\ln n}}$}
\textbf{解}\quad
$$
	\begin{aligned}
		\sum\limits_{n=1}^{\infty} \frac{1}{n 2^{\ln n}} & =\sum\limits_{n=1}^{\infty} \frac{1}{n e^{\ln 2 \ln n}} \\
		                                          & <\sum\limits_{n=1}^{\infty} \frac{1}{n^{\ln 4}}
	\end{aligned}
$$
收敛,由比较判别法知$\sum\limits_{n=1}^{\infty} \frac{1}{n 2^{\ln n}}$收敛.

\subsection{ $\sum\limits_{n=1}^{\infty} \frac{1}{n !}$}
\textbf{解}\quad
由$\mathrm{D'Alembert}$判别法:
$$
	\varlimsup_{n\rightarrow \infty}\frac{\frac{1}{(n+1)!}}{\frac{1}{n!}}=0<1
$$
收敛.

\subsection{ $\sum\limits_{n=1}^{\infty}\left(\frac{n^{2}}{2 n^{2}+1}\right)^{n}$}
\textbf{解}\quad
由柯西判别法:
$$
	\varlimsup_{n\rightarrow \infty}\sqrt[n]{\left(\frac{n^{2}}{2 n^{2}+1}\right)^{n}}=\varlimsup_{n\rightarrow \infty}\frac{n^{2}}{2n^{2}+1}=\frac{1}{2}<1
$$
收敛.

\subsection{ $\sum\limits_{n=2}^{\infty} \frac{1}{(\ln n)^{k}}(k>0)$}
\textbf{解}\quad
由柯西积分判别法:
$$
	\int_{n=2}^{\infty}\frac{1}{(\ln x)^{k}}\mathrm{~d}x>\int_{n=2}^{\infty}\frac{1}{x}\mathrm{~d}x=+\infty
$$
发散.

\subsection{ $\sum\limits_{n=1}^{\infty} \frac{1}{n^{1+\frac{1}{n}}}$}
\textbf{解}\quad
$$
	\begin{aligned}
		\lim\limits_{n \to \infty} \frac{\frac{1}{n^{1+\frac{1}{n}}}}{\frac{1}{n}}=\lim\limits_{n \to \infty}  \frac{1}{\sqrt[n]{n}}=1
	\end{aligned}
$$
由于$\sum\limits_{n=1}^{\infty}\frac{1}{n}$发散,从而$\sum\limits_{n=1}^{\infty} \frac{1}{n^{1+\frac{1}{n}}}$发散.

\subsection{ $\sum\limits_{n=1}^{\infty} \frac{1}{n} \sin \frac{1}{n}$}
\textbf{解}\quad
$$
	\begin{aligned}
		\sum\limits_{n=1}^{\infty} \frac{1}{n} \sin \frac{1}{n} & =\sum\limits_{n=1}^{\infty} \frac{1}{n} \left(\frac{1}{n}+O\left(\frac{1}{n^{3}}\right)\right)
	\end{aligned}
$$
显然$\sum\limits_{n=1}^{\infty}\frac{1}{n^{4}}$绝对收敛,从而由于$\sum\limits_{n=1}^{\infty}\frac{1}{n^{2}}$收敛,$\sum\limits_{n=1}^{\infty} \frac{1}{n} \sin \frac{1}{n}$收敛.

\subsection{ $\sum\limits_{n=1}^{\infty} \frac{1}{2^{\sqrt{n}}}$}
\textbf{解}\quad
由$\mathrm{Rabbe}$判别法,
$$
	\begin{aligned}
		\varliminf_{n \rightarrow \infty}n\left(\frac{\frac{1}{2^{\sqrt{n}}}}{\frac{1}{2^{\sqrt{n+1}}}}-1\right) & =\varliminf_{n \rightarrow \infty}n\left(2^{\frac{1}{\sqrt{n+1}+\sqrt{n}}}-1\right)=\varliminf_{n \rightarrow \infty}\frac{n}{\sqrt{n+1}+\sqrt{n}}=+\infty>1
	\end{aligned}
$$
收敛.

\subsection{ $\sum\limits_{n=1}^{\infty} n^{k} \mathrm{e}^{-n}$}
\textbf{解}\quad
由柯西判别法知:
$$
	\begin{aligned}
		\varlimsup_{n \to \infty} \sqrt[n]{n^{k} \mathrm{e}^{-n}}=\frac{1}{e}<1
	\end{aligned}
$$
收敛.

\subsection{ $\sum\limits_{n=1}^{\infty} \frac{(\ln n)^{k}}{n^{2}}$}
\textbf{解}\quad
$$
	\begin{aligned}
		\sum\limits_{n=1}^{\infty} \frac{(\ln n)^{k}}{n^{2}}<\sum\limits_{n=1}^{\infty} \frac{1}{n^{\frac{3}{2}}}
	\end{aligned}
$$
由于$\sum\limits_{n=1}^{\infty} \frac{1}{n^{\frac{3}{2}}}$收敛,由比较判别法知$\sum\limits_{n=1}^{\infty} \frac{(\ln n)^{k}}{n^{2}}$收敛.

\subsection{ $\sum\limits_{n=1}^{\infty} \frac{\left[2+(-1)^{n}\right]^{n}}{2^{2 n+1}}$}
\textbf{解}\quad
由柯西判别法知:
$$
	\begin{aligned}
		\varlimsup_{n \to \infty} \sqrt[n]{\frac{\left[2+(-1)^{n}\right]^{n}}{2^{2 n+1}}}=\frac{3}{4}<1
	\end{aligned}
$$
收敛.

\subsection{ $\sum\limits_{n=1}^{\infty}\left(\sqrt{n^{2}+1}-\sqrt{n^{2}-1}\right)$}
\textbf{解}\quad
$$
	\begin{aligned}
		\sum\limits_{n=1}^{\infty}\left(\sqrt{n^{2}+1}-\sqrt{n^{2}-1}\right)=\sum\limits_{n=1}^{\infty}\frac{2}{\sqrt{n^{2}+1}+\sqrt{n^{2}-1}}>\sum\limits_{n=1}^{\infty}\frac{1}{n}
	\end{aligned}
$$
由于$\sum\limits_{n=1}^{\infty}\frac{1}{n}$发散,故$\sum\limits_{n=1}^{\infty}\left(\sqrt{n^{2}+1}-\sqrt{n^{2}-1}\right)$发散.

\subsection{ $\sum\limits_{n=3}^{\infty}\left(-\ln \cos \frac{\pi}{n}\right)$}
\textbf{解}\quad
由
$$\cos \frac{\pi}{n}=1-\frac{\pi^{2}}{2 n^{2}}+o\left(\frac{\pi^{3}}{n^{3}}\right), n \rightarrow \infty$$
得
$$
	-\ln \cos \frac{\pi}{n}=-\ln \left[1-\frac{\pi^{2}}{2 n^{2}}+o\left(\frac{\pi^{3}}{n^{3}}\right)\right]=\frac{\pi^{2}}{2 n^{2}}+o\left(\frac{1}{n^{2}}\right) \sim \frac{1}{n^{2}}
$$
$$\sum\limits_{n=1}^{\infty} \frac{1}{n^{2}}$$ 收敛, 由比较判别法得: $$\sum\limits_{n=1}^{\infty}\left(-\ln \cos \frac{\pi}{n}\right)$$ 收敛

\subsection{ $\sum\limits_{n=1}^{\infty}(\sqrt[n]{n}-1)$}
\textbf{解}\quad
$$
	\begin{aligned}
		\sum\limits_{n=1}^{\infty}(\sqrt[n]{n}-1)=\sum\limits_{n=1}^{\infty}\left(e^{\frac{\ln n}{n}}-1\right)=\sum\limits_{n=1}^{\infty}\left(\frac{\ln n}{n}+O\left(\frac{\ln n}{n}\right)^{2}\right)
	\end{aligned}
$$
显然
$$
	\sum\limits_{n=1}^{\infty}\left(\frac{\ln n}{n}\right)^{2}
$$
绝对收敛,而
$$
	\sum\limits_{n=1}^{\infty}\frac{\ln n}{n}
$$
发散,从而
$$
	\sum\limits_{n=1}^{\infty}(\sqrt[n]{n}-1)
$$
发散.

\section{证明:若 $\lim _{n \rightarrow \infty} n x_{n}=a \neq 0,$ 则级数 $\sum\limits_{n=1}^{\infty} x_{n}$ 发散.}
\textbf{证}\quad
由于
$$
	\lim\limits_{n \to \infty} \frac{x_{n}}{\frac{1}{n}}=a
$$
从而级数与$\sum\limits_{n=1}^{\infty}\frac{1}{n}$同敛散.所以级数发散.证毕.

\section{证明:若正项级数 $\sum\limits_{n=1}^{\infty} x_{n}$ 收敛,则级数 $\sum\limits_{n=1}^{\infty} x_{n}^{2}$ 收敛. 举例说明反之不成立.}
\textbf{证}\quad
$$
	\lim\limits_{n \to \infty}  \frac{(x_{n})^{2}}{x_{n}}=\lim\limits_{n \to \infty}  x_{n}=0
$$
收敛.举例:$\{x_{n}\}=\frac{1}{n^{2}}$

\section{证明:若级数 $\sum\limits_{n=1}^{\infty} x_{n}^{2}, \sum\limits_{n=1}^{\infty} y_{n}^{2}$ 收敛,则级数 $\sum\limits_{n=1}^{\infty}\left(x_{n}+y_{n}\right)^{2}$ 和级数 $\sum\limits_{n=1}^{\infty}\left|x_{n} y_{n}\right|$ 都收敛.}
\textbf{证}\quad
由四则运算法则知$\sum\limits_{n=1}^{\infty}x_{n}^{2}+y_{n}^{2}$收敛.
$$
	2\sum\limits_{n=1}^{\infty}x_{n}^{2}+y_{n}^{2}\geqslant \sum\limits_{n=1}^{\infty}x_{n}^{2}+y_{n}^{2}+2|x_{n}y_{n}|\geqslant 4|x_{n}y_{n}|
$$
证毕.

\section{证明:若正项级数 $\sum x_{n}$ 收敛,则级数 $\sum\limits_{n=1}^{\infty} \frac{\sqrt{x_{n}}}{n^{p}}\left(p>\frac{1}{2}\right)$ 收敛.}
\textbf{证}\quad
$$
	\begin{aligned}
		\sum\limits_{n=1}^{\infty} \frac{\sqrt{x_{n}}}{n^{p}}\left(p>\frac{1}{2}\right)\leqslant \sum\limits_{n=1}^{\infty} \frac{x_{n}+\frac{1}{n^{2p}}}{2}
	\end{aligned}
$$
$\mathrm{RHS}$收敛,从而$\mathrm{LHS}$收敛.

\section{设 $\sum\limits_{n=1}^{\infty} x_{n}$ 是一个发散的正项级数,试证 $: \sum\limits_{n=1}^{\infty} \frac{x_{n}}{1+x_{n}}$ 发散.}
\textbf{证}\quad
若$\sum\limits_{n=1}^{\infty} \frac{x_{n}}{1+x_{n}}$收敛,则有$\lim\limits_{n \to \infty} \frac{x_{n}}{1+x_{n}}=\frac{1}{1+\frac{1}{x_{n}}}=0$,推知$\lim\limits_{n \to \infty} x_{n}=0$.\par 
由于数列极限存在且为$0$,故其存在最大项,记为$a_{k}=M$.故有:
$$
\sum\limits_{n=1}^{\infty} \frac{x_{n}}{1+M}\leqslant \sum\limits_{n=1}^{\infty} \frac{x_{n}}{1+x_{n}}
$$
前者发散,从而后者发散.证毕.

\section{设 $I_{n}=\int_{0}^{\frac{\pi}{4}} \tan ^{n} x \mathrm{~d} x, n=1,2, \cdots$}
\subsection{求级数 $\sum\limits_{n=1}^{\infty} \frac{I_{n}+I_{n+2}}{n}$ 的和}
\textbf{解}\quad
$$
	\frac{I_{n}+I_{n+2}}{n}=\frac{1}{n}\int_{0}^{\frac{\pi}{4}} \tan ^{n} x\sec^{2} x \mathrm{~d} x=\frac{1}{n}\int_{0}^{\frac{\pi}{4}} \tan ^{n} x \mathrm{~d} \tan x =\frac{1}{n(n+1)}
$$
从而
$$
	\sum\limits_{n=1}^{\infty} \frac{I_{n}+I_{n+2}}{n}=1
$$

\subsection{设 $\lambda>0,$ 证明级数 $\sum\limits_{n=1}^{\infty} \frac{I_{n}}{n^{\lambda}}$ 收敛.}
\textbf{证}\quad
$$
	\sum\limits_{n=1}^{\infty} \frac{I_{n}}{n^{\lambda}}<\sum\limits_{n=1}^{\infty} \frac{I_{n}+I_{n+2}}{n^{\lambda}}<\sum\limits_{n=1}^{\infty}\frac{1}{n^{(1+\lambda})}
$$
$\mathrm{RHS}$收敛,从而$\sum\limits_{n=1}^{\infty} \frac{I_{n}}{n^{\lambda}}$ 收敛.

\section{讨论下列级数的敛散性}
\subsection{ $\sum\limits_{n=1}^{\infty} n \tan \frac{\pi}{2^{n}}$}
\textbf{解}\quad
将$\tan \frac{\pi}{2^{n}}$泰勒展开:
$$
	\tan \frac{\pi}{2^{n}}=\frac{\pi}{2^{n}}+O\left(\frac{\pi^{3}}{2^{3n}}\right)
$$
显然对于级数$\sum\limits_{n=1}^{\infty} \frac{n}{2^{3n}}$收敛(由柯西判别法可易得),又因为$\sum\limits_{n=1}^{\infty}\frac{n}{2^{n}}$
收敛(同上理),所以级数收敛.

\subsection{ $\sum\limits_{n=1}^{\infty} \frac{n^{k}}{3^{n}}$}
\textbf{解}\quad
由柯西判别法知:
$$
	\varlimsup_{n \to \infty} \sqrt[n]{\frac{n^{k}}{3^{n}}}=\frac{1}{3}<1
$$
从而级数收敛.

\subsection{ $\sum\limits_{n=1}^{\infty} \frac{n^{2}}{\left(2+\frac{1}{n}\right)^{n}}$}
\textbf{解}\quad
由柯西判别法知:
$$
	\varlimsup_{n \to \infty} \sqrt[n]{\frac{n^{2}}{\left(2+\frac{1}{n}\right)^{n}}}=\frac{1}{2}<1
$$
从而级数收敛.

\subsection{ $\sum\limits_{n=1}^{\infty} \frac{n^{n+\frac{1}{n}}}{\left(n+\frac{1}{n}\right)^{n}}$}
\textbf{解}\quad
$$
	\lim\limits_{n \to \infty} \frac{n^{n+\frac{1}{n}}}{\left(n+\frac{1}{n}\right)^{n}}=1\neq 0
$$
发散.

\subsection{ $\sum\limits_{n=2}^{\infty} \frac{n^{\ln n}}{(\ln n)^{n}}$}
\textbf{解}$1^{\circ}$\quad
由柯西判别法:
$$
	\begin{aligned}
		\varlimsup_{n \to \infty}\sqrt[n]{\frac{n^{\ln n}}{(\ln n)^{n}}} & =\varlimsup_{n \to \infty}\frac{n^{\frac{\ln n}{n}}}{\ln n}=\varlimsup_{x \to \infty}\frac{x^{\frac{\ln x}{x}}}{\ln x} \\
		                                                                 & =\varlimsup_{x \to \infty}\frac{x^{\frac{\ln x}{x}}(2\ln x -\ln^{2}x)}{x}                                              \\
		                                                                 & =\varlimsup_{x \to \infty}\frac{x^{\frac{\ln x}{x}}(2\ln x - \ln^{2}x)^{2}}{x^{2}}+x^{\frac{\ln x - x}{x}}(2-2\ln x)   \\
		                                                                 & =0
	\end{aligned}
$$
收敛.

\textbf{解}$2^{\circ}$\quad
因为$$a_{n}=\frac{n^{\ln n}}{(\ln n)^{n}}=e^{\ln ^{2} n-n \ln (\ln n)}=e^{-\left(n \ln \ln n-\ln ^{2} n\right)}$$
由于 $$\lim _{n \rightarrow \infty} \frac{n \ln \ln n-\ln ^{2} n}{n}= +\infty$$
所以$\exists N,$ 当 $n>N$ 时,有 $n \ln \ln n-\ln ^{2} n \geqslant A \cdot n,$ 其中 $A>0 .$ 从而
$$\frac{a_{n}}{\frac{1}{n^{2}}}=\frac{n^{2}}{e^{n \ln \ln n-\ln ^{2} n}} \leqslant \frac{n^{2}}{e^{A n}} \rightarrow 0(n \rightarrow \infty)$$
于是原级数收敛.

\subsection{ $\sum\limits_{n=1}^{\infty} \frac{2+(-1)^{n}}{2^{n}}$}
\textbf{解}\quad
由柯西判别法:
$$
	\varlimsup_{n \to \infty} \sqrt[n]{\frac{2+(-1)^{n}}{2^{n}}}=\frac{1}{2}<1
$$
收敛.

\section{设 $x_{n}>0, \frac{x_{n+1}}{x_{n}}>1-\frac{1}{n}(n=1,2, \cdots)$. 证明:级数 $\sum\limits_{n=1}^{\infty} x_{n}$ 发散.}
\textbf{证}$1^{\circ}$\quad
由$\mathrm{Bertrand}$判别法:
$$
	\varlimsup_{n \to \infty} \ln n\left[n\left(\frac{x_{n}}{x_{n+1}}-1\right)-1\right]=\varlimsup_{n \to \infty}\frac{\ln n}{n-1}=0<1
$$
发散.

\textbf{证}$2^{\circ}$\quad
$$
	\left.\begin{array}{l}
		\frac{x_{n+1}}{x_{n}}>1-\frac{1}{n}=\frac{n-1}{n} \\
		\frac{x_{n}}{x_{n-1}}>\frac{n-2}{n-1}             \\
		\ldots \ldots \ldots                              \\
		\frac{x_{3}}{x_{2}}>\frac{1}{2}
	\end{array}\right\} \Rightarrow \frac{x_{n+1}}{x_{2}}>\frac{2}{n}, x_{n+1}>\frac{2 x_{2}}{n}\sim \frac{1}{n}
$$
由于级数$\sum\limits_{n=1}^{\infty}\frac{1}{n}$发散,从而原级数发散.

注:关于其它判别法和对于已知判别法的扩展可以参考“学术启航”公众号中的工科数分下册的理论知识模块中的项级数判别法栏,或者直接回复判别法.

\section{讨论下列级数的敛散性}
\subsection{ $\sum\limits_{n=1}^{\infty} \frac{100^{n}}{n !}$}
\textbf{解}\quad
由$\mathrm{D'Alembert}$判别法:
$$
	\varlimsup_{n \to \infty} \frac{x_{n+1}}{x_{n}}=\varlimsup_{n \to \infty}\frac{\frac{100^{n+1}}{(n+1)!}}{\frac{100^{n}}{n!}}=\varlimsup_{n \to \infty}\frac{100}{(n+1)}=0<1
$$
收敛.

\subsection{ $\sum\limits_{n=1}^{\infty} \frac{(n !)^{2}}{(2 n) !}$}
\textbf{解}\quad
由$\mathrm{D'Alembert}$判别法:
$$
	\varlimsup_{n \to \infty}\frac{x_{n+1}}{x_{n}}=\varlimsup_{n \to \infty}\frac{\frac{[(n+1)!]^{2}}{(2n+2)!}}{\frac{(n!)^{2}}{(2n)!}}=\varlimsup_{n \to \infty}\frac{(n+1)^{2}}{(2n+2)(2n+1)}=\frac{1}{4}<1
$$
收敛.

\subsection{ $\sum\limits_{n=1}^{\infty} \frac{n !}{n^{n}}$}
\textbf{解}\quad
由$\mathrm{D'Alembert}$判别法:
$$
	\varlimsup_{n \to \infty}\frac{x_{n+1}}{x_{n}}=\varlimsup_{n \to \infty}\frac{\frac{(n+1)!}{(n+1)^{(n+1)}}}{\frac{n!}{n^{n}}}=\varlimsup_{n \to \infty}\frac{1}{\left(1+\frac{1}{n}\right)^{n}}=\frac{1}{e}<1
$$
收敛.

\subsection{ $\sum\limits_{n=1}^{\infty} \frac{2^{n} n !}{n^{n}}$}
\textbf{解}\quad
由$\mathrm{D'Alembert}$判别法:
$$
	\varlimsup_{n \to \infty}\frac{x_{n+1}}{x_{n}}=\frac{2}{e}<1
$$
收敛.


\subsection{ $\sum\limits_{n=1}^{\infty} \frac{n^{n}}{(n !)^{2}}$}
\textbf{解}\quad
由$\mathrm{D'Alembert}$判别法:
$$
	\varlimsup_{n \to \infty}\frac{x_{n+1}}{x_{n}}=\varlimsup_{n \to \infty}\frac{1}{e(n+1)}=0<1
$$
收敛.



\subsection{ $\sum\limits_{n=1}^{\infty} \frac{(2 n) !}{2^{n(n+1)}}$}
\textbf{解}\quad
由$\mathrm{D'Alembert}$判别法:
$$
\varlimsup_{n \to \infty}\frac{x_{n+1}}{x_{n}}=\varlimsup_{n \to \infty}\frac{(2n+2)(2n+1)}{4^{n+1}}=0<1
$$
发散.

\section{已知 $f(x)$ 在区间 [1, $+\infty$ ) 上二阶可导,且 $f^{\prime \prime}(x)<0(\forall x \in(0,+\infty)), \lim _{x \rightarrow+\infty} f(x)=$
$A .$ 证明:级数 $\sum\limits_{n=1}^{\infty} f^{\prime}(n)$ 收敛.}
\textbf{证}$1^{\circ}$\quad
由$f(x)$二阶可导且其二阶导数恒负,得$f'(x)$连续严格单减.由于$f(x)$极限为$A$.已知$\forall \varepsilon>0$,存在$N$使得满足$\forall x>N$,$|f(x)-A|<\varepsilon$.从而取$x',x'+1>N+1$,则由$\mathrm{Lagrange}$中值定理可知$\exists \xi \in (x',x'+1)$,$|f'(\xi)|<\varepsilon$.从而$f'(x)$极限为$0$.

由柯西积分判别法知$\sum\limits_{n=1}^{\infty} f^{\prime}(n)$与$\int_{1}^{\infty}f'(x)\mathrm{~d}x$同敛散.又因为
$$
\int_{1}^{\infty}f'(x)\mathrm{~d}x=A-f(1)
$$
从而级数收敛.

\textbf{证}$2^{\circ}$\quad
由$\mathrm{Lagrange}$ 中值定理, $\quad \exists \xi \in(n-1, n)\mathrm{s.t.}f^{\prime}(\xi)=\frac{f(n)-f(n-1)}{n-(n-1)}$,且 $f^{\prime \prime}(n)<0$
则 
$$
f^{\prime}(n)<f^{\prime}(\xi)=f(n)-f(n-1)
$$
因为 
$$
\sum\limits_{n=1}^{\infty}[f(n)-f(n-1)]=\lim _{n \rightarrow \infty}[f(n)-f(0)]=A-f(0)
$$
而 
$$
\sum\limits_{n=1}^{\infty} f^{\prime}(n)<\sum\limits_{n=1}^{\infty}[f(n)-f(n-1)]
$$ 
则级数 $\sum\limits_{n=1}^{\infty} f^{\prime}(n)$ 收敛.

\section{讨论下列级数的敛散性}
\subsection{$\sum\limits_{n=2}^{\infty} \frac{1}{\ln (n !)}$}
\textbf{解}\quad
由于
$$
\sum\limits_{n=2}^{\infty} \frac{1}{\ln (n !)}>\sum\limits_{n=2}^{\infty} \frac{1}{n\ln n}
$$
$\mathrm{RHS}$发散,从而原级数发散.

\subsection{$\sum\limits_{n=1}^{\infty} \frac{1}{\ln (1+n)} \sin \frac{1}{n}$}
\textbf{解}\quad
由于
$$
\sin \frac{1}{n}=\frac{1}{n}+O\left(\frac{1}{n^{3}}\right)
$$
又因为$\sum\limits_{n=1}^{\infty}\frac{1}{n^{3}\ln(1+n)}$绝对收敛,$\sum\limits_{n=1}^{\infty}\frac{1}{n\ln (n+1)}$发散,从而原级数发散.

\subsection{$\sum\limits_{n=2}^{\infty} \frac{1}{n \ln n \ln \ln n}$}
\textbf{解}$1^{\circ}$\quad
存在函数$f(x)=\frac{1}{x\ln x\ln \ln x}$连续单减趋于$0$,由柯西积分判别法知:
$$
\int_{2}^{\infty}\frac{1}{x\ln x\ln \ln x}\mathrm{~d}x=\left.\ln \ln x\right|_{2}^{+\infty}=+\infty
$$
发散.

\textbf{解}$2^{\circ}$\quad
由$\mathrm{Bertrand}$判别法:
$$
\begin{aligned}
    \lim\limits_{n \to \infty}\ln \ln \ln n\left\lbrace\ln \ln n\left\lbrace \ln n \left[n\left(\frac{x_{n}}{x_{n+1}}-1\right)-1\right]-1\right\rbrace-1\right\rbrace&=0<1
\end{aligned}
$$
发散.

可以利用$\mathrm{Matlab}$进行验证:

>>syms x 

>>f =log(log(log(x)))*(log(log(x))*(log(x)*(x*((log(x + 1)*log(log(x + 1))*(x + 1))/(x*log(log(x))*log(x)) - 1) - 1) - 1) - 1)

>>limit(f,x,inf)

得到结果:

>>ans=0


\section{使用 $\mathrm{Raabe}$ 判别法判断下列级数的敛散性:}
\subsection{ $\sum\limits_{n=1}^{\infty} \frac{n !}{(a+1)(a+2) \cdots(a+n)}(a>0)$}
\textbf{解}\quad
$$
\begin{aligned}
    \lim\limits_{n \to \infty}n\left(\frac{x_{n}}{x_{n+1}}-1\right)&=\lim\limits_{n \to \infty}\frac{a\cdot n}{n+1}
\end{aligned}
$$
当$a>1$时收敛,当$0<a<1$时发散.当$a=1$时,$\sum\limits_{n=1}^{\infty} \frac{n !}{(a+1)(a+2) \cdots(a+n)}=\sum\limits_{n=1}^{\infty} \frac{1}{n+1}$发散.



\subsection{ $\sum\limits_{n=1}^{\infty}\left(\frac{1}{2}\right)^{1+\frac{1}{2}+\cdots+\frac{1}{n}}$}
\textbf{解}\quad
$$
\begin{aligned}
    \varlimsup_{n \to \infty}n\left(\frac{x_{n}}{x_{n+1}}-1\right)&=\varlimsup_{n \to \infty}n\left(2^{\frac{1}{n+1}}-1\right)=\varlimsup_{n \to \infty} \frac{n\ln 2}{n+1}=\ln 2<1
\end{aligned}
$$
发散.


\subsection{ $\sum\limits_{n=1}^{\infty} \frac{\sqrt{n !}}{(a+\sqrt{1})(a+\sqrt{2}) \cdots(a+\sqrt{n})}(a>0)$}
\textbf{解}\quad
$$
\begin{aligned}
    \varliminf_{n \to \infty}n\left(\frac{x_{n}}{x_{n+1}}-1\right)&=\varliminf_{n \to \infty}n\left(\frac{a+\sqrt{n+1}}{\sqrt{n+1}}-1\right)=\varliminf_{n \to \infty}\frac{a\cdot n}{\sqrt{n+1}}=+\infty>1
\end{aligned}
$$
收敛.

\section{设正项级数 $\sum\limits_{n=1}^{\infty} x_{n}\left(x_{n}>0, n=1,2, \cdots\right)$ 发散,证明:存在一个发散的正项级数 $\sum\limits_{n=1}^{\infty} y_{n}$,使得 $\lim _{n \rightarrow \infty} \frac{y_{n}}{x_{n}}=0$}
\textbf{证}$1^{\circ}$\quad
由正项级数 $\sum\limits_{n=1}^{\infty} a_{n}$ 收敛知，余项 $R_{n}$ 单调减少收敛于 $0 .$ 令 $b_{n}=$ $\sqrt{R_{n-1}}-\sqrt{R_{n}},$ 其中记 $R_{0}=\sum\limits_{n=1}^{\infty} a_{n} .$从
$$
\frac{a_{n}}{b_{n}}=\frac{R_{n-1}-R_{n}}{\sqrt{R_{n-1}}-\sqrt{R_{n}}}=\sqrt{R_{n-1}}+\sqrt{R_{n}} \rightarrow 0
$$
和 $\sum\limits_{k=1}^{n} b_{k}=\sqrt{R_{0}}-\sqrt{R_{k}} \leqslant \sqrt{R_{0}}$ 可见 $\sum\limits_{n=1}^{\infty} b_{n}$ 为满足要求的收敛级数.

\textbf{证}$2^{\circ}$\quad
由$\mathrm{Sapagof}$判别法知$\sum\limits_{n=1}^{\infty}a_{n}$与$\sum\limits_{n=1}^{\infty}\frac{a_{n}}{S_{n}}$同敛散($S_{n}$是$a_{n}$的部分和序列)从而令$y_{n}=\frac{x_{n}}{S_{n}}$即可.

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